the vertex is at (214.26,102.51)
So, y = C(x-214.26)^2 + k
102.51 = C(0)^2 + k
so, k = 102.51
y = C(x-214.26)^2 + 102.51
0 = C(432-214.26)^2 + 102.51
C = -0.002162
y = -0.002162(x-214.26)^2 + 102.51
y = -.002162x^2 + .92646x + 3.25831
y = -.002162(x+3.4885)(x-432)
Can you please help me with a math problem
Suppose the ball was 3.24 feet above ground when it was hit, and that it reached a maximum height of approximately 102.51 feet when it had traveled a ground distance of approximately 214.26 feet. The ball lands after traveling a ground distance of approximately 432 feet.
Find an equation of the form y = C(x-z1)(x-z2) where z1 and z2 are the zeros (or roots) of the quadratic polynomial (or x-intercepts of the graph) and C is a scaling constant that needs to be determined.
Thanks!
2 answers
Why are these poor models for a parabola, Where a ball starts a a certain point and then is hit to reach a maximum height (vertex) and then lands at a certain point
i. y = -0.002x(x - 437.1)
ii. y = -0.5x + 216x + 3
iii. y = -0.002x + 0.879x + 3.981
iv. y = -0.002x + 0.8732x - 3.981
Thanks!
i. y = -0.002x(x - 437.1)
ii. y = -0.5x + 216x + 3
iii. y = -0.002x + 0.879x + 3.981
iv. y = -0.002x + 0.8732x - 3.981
Thanks!